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2 years ago

Helppppppp pleaseee

Mathematics

1 answer:

Viktor [21]2 years ago

5 0

Answer:

$x=2\sqrt{5}$

Step-by-step explanation:

If that is a square, then all sides are the same. Therefore, it's a $\sqrt{10}$ by $\sqrt{10}$ shape. We'll use Pythagorean Theorem to find the diagonal.

$a^2+b^2=c^2\\\sqrt{10}^2+ \sqrt{10}^2=x^2\\10+10=x^2\\20=x^2\\\sqrt{20}=x\\\sqrt{4*5}=x\\ 2\sqrt{5}=x$

Since it's not a fraction, there is no denominator to rationalize. Let me know if you have any other questions!

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Please help!! x = 2 - 3 cos t, y = 1 + 4 sin t in rectangular form?<br><br> thank you! :)

sukhopar [10]

Answer:

$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1$

Step-by-step explanation:

$x=2-3\cos(t)$

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Let's solve for $\cos(t)$ in the first equation and then solve for $\sin(t)$ in the second equation.

I will then use the following identity to get right of the parameter, $t$ :

$\cos^2(t)+\sin^2(t)=1$ (Pythagorean Identity).

Let's begin with $x=2-3\cos(t)$ .

Subtract 2 on both sides:

$x-2=-3\cos(t)$

Divide both sides by -3:

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Subtract 1 on both sides:

$y-1=4\sin(t)$

Divide both sides by 4:

$\frac{y-1}{4}=\sin(t)$

Now let's plug it into our Pythagorean Identity:

$\cos^2(t)+\sin^2(t)=1$

$\frac{x-2}{-3})^2+(\frac{y-1}{4})^2=1$

$\frac{(x-2)^2}{(-3)^2}+\frac{(y-1)^2}{4^2}=1$

$\frac{(x-2)^2}{9}+\frac{(y-1)^2}{16}=1$

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2 years ago

Read 2 more answers

Help with math please example 1

hram777 [196]

Answer:

I'm not very sure about this one what icing in probably not going to be right but what I think is that they're not the same because look at the size difference imagine putting an adult's next to a baby that's kind of what it's like the bigger quadrilateral is also the adults while the little one is the baby what do you think about that I probably think that they're not saying

Step-by-step explanation:

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tatuchka [14]

Answer would be C

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olasank [31]

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Why does these expressions 2(30x-24), 3(20x-16), and 4(15x-12) have the same expansion?

MariettaO [177]

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